Question

Answer the following questions showing all work. Full credit will not be given to answers without work shown. If you use Minitab Express or StatKey include the appropriate output (copy + paste). If you do hand calculations show your work using the Word equation editor. Clearly identify your final answers. Output without explanation will not receive full credit and answers with no output or explanation will not receive full credit. Round all answers to 3 decimal places. If you have any questions, post them to the course discussion board.

Part 1: Two Independent Proportions
Reminder: The standard error is computed differently for a two sample proportion confidence interval and a two sample proportion hypothesis test.

In a random sample of 35 Penn State Shenango students, 17 were attending full-time. In a random sample of 40 World Campus students, 13 were attending full-time.

A.   Use Minitab Express to construct a 95% confidence interval to estimate the difference between the proportions of all Penn State Shenango and all Penn State World Campus students who are attending full-time. If assumptions are met, use the normal approximation method. Do not do any calculations by hand. Remember to copy+paste all relevant Minitab Express output. Always clearly identify your final answer. [20 points]

B. Use Minitab Express and the five-step hypothesis testing procedure given below to determine if there is evidence of a difference between the proportions of all Penn State Shenango and all Penn State World Campus students who are attending full-time. If assumptions are met, use the normal approximation method. Do not do any calculations by hand. Remember to copy+paste all relevant Minitab Express output. [30 points]
Step 1: Check assumptions and write hypotheses
Step 2: Calculate the test statistic
Step 3: Determine the p-value
Step 4: Decide between the null and alternative hypotheses
Step 5: State a real world conclusion

Part 2: Two Independent Means
Data were collected from a random sample of American Facebook accounts. Data concerning gender and number of Facebook friends are in the file FacebookGender.MTW.

A. Construct side-by-side box plots to compare men and women in terms of the number of Facebook friends they have. [10 points]

B. Construct a dotplot with groups to compare men and women in terms of the number of Facebook friends they have. [10 points]

C. Use Minitab Express and the five-step hypothesis testing procedure given below to determine if American men and women differ in the population in terms of the number of friends they have on Facebook on average. If assumptions are met, use a t distribution to approximate the sampling distribution. Do not do any calculations by hand. Remember to include all relevant Minitab Express output. [30 points]
Step 1: Check assumptions and write hypotheses
Step 2: Calculate the test statistic
Step 3: Determine the p-value
Step 4: Decide between the null and alternative hypotheses
Step 5: State a real world conclusion

Answer 1

1.A)

• Check assumptions and write hypotheses​ :

Let Xi = scores of lesson 5 quiz of ith student ~ Normal ($\mu_1,\sigma^2$) and

Yi=  scores of lesson 4 quiz of ith student ~ Normal ($\mu_2,\sigma^2$)

Then di=Xi-Yi = Lesson 5 Quiz – Lesson 4 Quiz ~ N($\mu_1-\mu_2,\sigma^2$ )

hypothesis : Ho : $\mu_1= \mu_2$ vs H1: $\mu_1> \mu_2$

• Calculate the test statistic:

Test statistic(T) =
SE(d)

tin_

=0.2935 /1.8003 = 0.163

• Determine P value :

p value = Prob( |Ho is true) = 0.178

T ta/2

• Decide between Null and Alternative Hypothesis

As P value =0.178 >0.05 =size hence we dont reject the Null hypothesis Ho: u1-u2 = 0

• Interpretation :

the conclusion is that, statistically, the mean difference between the marks of week 5 quiz and week 4 quiz arent significantly different from zero.

B)

Minitab-Untitled - [Session] 雹File Edit Data Calc Stat Graph Editor Tools Window 브elp Assistant 10/24/2017 2:55:18 AM helcone to Minitab, preaa l for help Paired T-Test and CI N Mean StDew SE Mean Difference 33 0.293 1.800 0.313 95% lower bound for mean difference: -0.237 I-Iest of rean difference One-Sample T Test of m0 vs >0 (> 0) T-Value-0.94 P-Value- 0.178 95% Lower Bound N Mean StDew SE Mean 33 0.293 1.800 0.313 -0.237 0.94 0.178 Editable 3:11 AM Current Worksheet: Worksheet 1 O Type here to search ^ 41) ENG 10/24/2017

C)Why same ?

The answers in part a amd b are same because in both cases it is test of population mean(equal to zero) which is estimated by test statistic : sample mean over its standard error

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